A block of mass $M = 2 k g$ with a semicircular track of radius $R = 1.1 m$ rests on a horizontal frictionless surface. A uniform cylinder of radius $r = 10 c m$ and mass $m = 1.0 k g$ is released from rest from the top point $A$ . The cylinder slips on the semicircular frictionless track. The speed of the block when the cylinder reaches the bottom of the track at B is $(g = 10 m / s^{2})$

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$\sqrt{\frac{10}{3}} m / s$

$\sqrt{\frac{5}{2}} m / s$

$\sqrt{\frac{4}{3}} m / s$

$\sqrt{10} m / s$

answer is A.

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Detailed Solution

Let speed of block is v. Then from conservation of linear momentum in horizontal direction velocity of cylinder will by 2v in opposite direction $(a s m = \frac{M}{2})$ .
Now from conservation of mechanical energy we have
$m g h = \frac{1}{2} M v^{2} + \frac{1}{2} m {(2 v)}^{2}$
Here $h = R - r = 1.0 m$
Substituting the values, we get,
$(1) (10) (1) = \frac{1}{2} (2) (v^{2}) + \frac{1}{2} (1) (4 v^{2}) o r 3 v^{2} = 10 ∴ v = \sqrt{\frac{10}{3}} m / s$

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